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Ede4ka [16]
2 years ago
15

A uniform electric field of strength E points to the right. An electron is fired with a velocity v0 to the right and travels a d

istance d before coming to a stop. An second electron is then fired upwards through the same field at a velocity of v0. After the electron moving vertical has traveled vertically upwards a distance d, how far will it have moved horizontally?
Physics
1 answer:
zvonat [6]2 years ago
5 0

Answer:

<u />D_l=d<u />

Explanation:

From the question we are told that:

The Electric field of strength direction =Right

The Velocity of The First Electron=V_0

The Velocity of The Second Electron=V_0

Therefore

V_{e1}=V_{e2}

Generally, the equation for the Horizontal Displacement of electron is mathematically given by

D=\frac{at^2}{2}

Where

Acceleration is given as

a=\frac{V_o}{2d}

And

Time

T=\frac{d}{v_0}

Therefore horizontal displacement towards the left is

D_l=\frac{(\frac{V_o}{2d})(\frac{d}{v_0})^2}{2}

<u />D_l=d<u />

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You toss a 0.40-kg ball at 9.0 ms/ to a 14-kg dog standing on an iced-over pond. The dog catches the ball and begins to slide on
just olya [345]

Answer:

a)   v_{f} = 0.25 m / s  b) u = 0.25 m / s

Explanation:

a) To solve this problem let's start with the conservation of the moment, for this we define a system formed by the ball plus the dog, in this case all the forces are internal and the moment is conserved

We will write the data

     m₁ = 0.40 kg

     v₁₀ = 9.0 m / s

     m₂ = 14 kg

     v₂₀ = 0

Initial

     po = m₁ v₁₀

Final

     p_{f} = (m₁ + m₂) vf

     po = pf

     m₁ v₁₀ = (m₁ + m₂) v_{f}

      v_{f} = v₁₀ m₁ / (m₁ + m₂)

      v_{f} = 9.0 (0.40 / (0.40 +14)

      v_{f} = 0.25 m / s

b) This is the reference frame of the center of mass of the system in this case the speed of this frame is the speed of the center of mass

      u = 0.25 m / s

In the direction of movement of the ball

c) Let's calculate the kinetic energy in both moments

Initial

     K₀ = ½ m₁ v₁₀² +0

     K₀ = ½ 0.40 9 2

     K₀ = 16.2 J

Final

     K_{f}= ½ (m₁ + m₂) v_{f}2

      K_{f} = ½ (0.4 +14) 0.25 2

    K_{f} = 0.45 J

   

    ΔK = K₀ -  K_{f}

    ΔK = 16.2-0.445

    ΔK = 1575 J

These will transform internal system energy

d) In order to find the kinetic energy, we must first find the velocities of the individual in this reference system.

      v₁₀’= v₁₀ -u

      v₁₀’= 9 -.025

      v₁₀‘= 8.75 m / s

      v₂₀ ‘= v₂₀ -u

      v₂₀‘= - 0.25 m / s

      v_{f} ‘=   v_{f} - u

      v_{f} = 0

Initial

    K₀ = ½ m₁ v₁₀‘² + ½ m₂ v₂₀‘²

    Ko = ½ 0.4 8.75² + ½ 14.0 0.25²

    Ko = 15.31 + 0.4375

    K o = 15.75 J

Final

   k_{f} = ½ (m₁ + m₂) vf’²

  k_{f} = 0

All initial kinetic energy is transformed into internal energy in this reference system

3 0
2 years ago
Which is an example of a vector quantity
swat32

Answer:

The answer to your question is C

Explanation:

acceleration is a vector quantity because it has both magnitude and direction

4 0
3 years ago
A biker pedals hard to ride his bike to the top of a 44 m hill. He decides to let his bike coast down the hill, and is having so
Dafna11 [192]

Answer:

The bikers speed at the top of other hill is <u>25.82 m/s.</u>

Explanation:

Considering the biker is riding on a frictionless surface.

∴ There is no non-conservative or external force acting on the biker.

Hence we can conserve the energy of biker and bike as a system.

Let,

h_{1} = 44m

h_{2} = 10m

Since the biker starts from rest , his initial speed v_{1} = 0 m/s

Let final speed of the bike at the top of other hill be v_{2}.

∴ Initial Energy (at the top of 44m hill) = mgh_{1}

  Final Energy  (at the top of 10m hill) =  mgh_{2} + \frac{1}{2}mv_{2} ^{2}.

Conserving both the energies , we get

mgh_{1} = mgh_{2} + \frac{1}{2}mv_{2} ^{2}

∴ v_{2} = \sqrt{2g(h_{1}-h_{2} )}

Substituting the values for g , h_{1} , h_{2} , we get

v_{2} = 25.82 m/s

6 0
3 years ago
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Answer:

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Explanation:

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4 0
2 years ago
What is the rocks potential energy when it is 2000 meters above the ground?
zhuklara [117]

Answer:  

PE = 2032 J

m = 20 kg

g = 2 in/s2

h = 2000 m

8 0
3 years ago
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